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If V={(x,y)|x,y ∈ R} with acts (x1,y1)@(x2,y2)=(3(x13+x23),3(y13+y23)) and a$(x1,y1)=(ax1,ay1) , a ∈ R.

Prove V with this acts is NOT vector space.Thank you! 

 Nov 12, 2018
 #1
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are you sure it's not?

 

it's addition has commutativity and associativity

 

(0,0) is the clear additive identity and (-x,-y) is the additive inverse of (x,y)

 

the scalar properties are identical to the Euclidean vector space

 

I'm not seeing why this isn't a vector space.

 

remember the cube and cube root form a bijection

 Nov 12, 2018
 #2
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This is not a vector space. Exactly one of the axioms of a vector space is not satisfied.

Guest Nov 12, 2018
 #3
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Rom yes im sure the exercise say it's not.For this reason i can't solve the exercise

Dimitristhym  Nov 12, 2018
 #4
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Guest "Exactly one of the axioms of a vector space is not satisfied"

Witch is this axiom? 

Dimitristhym  Nov 12, 2018
 #5
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Ok.  Distribution of scalar mutliplication with respect to field addition doesn't hold.

 

we should have

 

(a+b)v=avbv

 

let's take a look

 

(a+b)v={(a+b)vx,(a+b)vy}avbv={3a3v3x+b3v3x,3a3v3y+b3v3y}={3a3+b3vx, 3a3+b3vy}3a3+b3a+b

Rom  Nov 13, 2018
 #6
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Thank you very much Rom! 

Dimitristhym  Nov 13, 2018

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